A small object placed on a rotating horizontal turn table just slips when it is placed at a distance 4 cm from the axis of rotation. If the angular velocity of the turn-table is doubled, the object slips when its distance from the axis of rotation is 

The object will slip if the required centripetal force $\ge$ force of friction 

                    ${\mathrm{mr\omega }}^2 \ge$$\mathrm{\mu mg}$

                   ${\mathrm{r\omega }}^2 \ge \mathrm{\mu g}$

                    ${\mathrm{r\omega }}^2 \ge \mathrm{constant}$

or $\left(\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}\right) = {\left(\frac{{\mathrm{\omega }}_2}{{\mathrm{\omega }}_1}\right)}^2$

$\frac{4 \mathrm{cm}}{{\mathrm{r}}_2} = {\left(\frac{2\mathrm{\omega }}{\mathrm{\omega }}\right)}^2$

${\mathrm{r}}_2= 1 \mathrm{cm}$